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Prelim problem: bizarre integral expression

  1. Jan 8, 2012 #1
    Hi all,
    Let [itex]f:[0,\infty)\rightarrow\mathbb{R}[/itex] be a bounded measurable function such that
    [tex]\lim_{x\rightarrow\infty}x^2f(x)=1.[/tex]
    Find an integral expression for
    [tex]\lim_{\lambda\rightarrow 0^+}\frac{\int_0^{\infty}(1-\cos(x))f(\frac{x}{\lambda})dx}{\lambda^2}.[/tex]

    This one is really bizarre to me. I'm not sure how to use the information about f's endpoint behavior other than to try somehow to approximate f by 1/x^2?

    I've tried changing lambda into 1/n where n goes to infinity but that seems to get me an expression that will tend to 0, not an integral expression as requested.

    Any help or ideas would be greatly appreciated; a solution isn't necessary, but maybe thoughts on how to think about the problem?

    Thanks!
     
  2. jcsd
  3. Jan 9, 2012 #2

    micromass

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    The theorem on dominant convergence??
     
  4. Jan 9, 2012 #3

    Stephen Tashi

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    [tex]\lim_{x\rightarrow\infty}x^2f(x)=1.[/tex]

    For k > 0, a constant

    [itex] \lim_{x \rightarrow \infty} (kx)^2 f(kx) = 1 [/itex]

    [itex] k^2 \lim_{x \rightarrow \infty} x^2 f(kx) = 1 [/itex]

    [itex] \lim_{x \rightarrow \infty} x^2 f(kx) = \frac{1}{k^2} [/itex]

    Change variables [itex] x = 1/ \lambda [/itex] and let the [itex] k [/itex] in the above equation be represented by [itex] x [/itex]

    [itex] \lim_{\lambda \rightarrow 0^+} \frac{1}{\lambda^2} f(\frac{x}{\lambda}) = \frac{1}{x^2}[/itex]

    Then we need an excuse (dominated convergence?) to take the limit inside the integral sign, in order to do the manipulations:

    [tex]\lim_{\lambda\rightarrow 0^+}\frac{\int_0^{\infty}(1-\cos(x))f(\frac{x}{\lambda})dx}{\lambda^2} [/tex]

    =

    [tex] \int_{0}^{\infty} (1 - cos(x)) \lim_{\lambda \rightarrow 0^+} \frac{1}{\lambda^2} f(\frac{x}{\lambda}) dx [/tex]

    =

    [tex] \int_{0}^{\infty} \frac{ (1 - cos(x))}{x^2} dx [/tex]
     
    Last edited: Jan 9, 2012
  5. Jan 9, 2012 #4
    Thank you for the insight, it was very helpful!
    To apply dominated convergence I think we have to be careful because f isn't integrable, but I think we can say something like
    [tex]\int_a^{\infty} f \leq \sum_i \int_{a_i}^{a_{i+1}}\frac{1}{x^2}+\frac{\epsilon}{2^i}\,dx<\infty.[/tex]
     
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